Describing Motion Around Us Class 9 Notes, Graphs and Numericals

What Describing Motion Around Us is about

Weightage: This chapter is part of Unit III, Motion, Force, Work and Sound, which carries 23 of the 80 theory marks in the school annual exam, shared with How Forces Affect Motion, Work, Energy, and Simple Machines and Sound Waves: Characteristics and Applications.Source: CBSE Science (086) syllabus, Class IX 2026-27, course structure, Unit III Motion, Force, Work and Sound

Key concepts

Position, distance travelled and displacement

NCERT section 4.1 Motion in a Straight Line

To say where something is, we need a fixed reference point (origin) and a direction. On a straight line, positions on one side are taken as positive (+) and on the other side as negative (−). An object is in motion if its position changes with time, and at rest if it does not.

• Distance travelled: the total length of the path covered. It has only a size (magnitude), never a direction, and never decreases.
• Displacement: the net change in position, from the first position to the final one. It has a magnitude and a direction.
• SI unit of both: the metre (m).

Key facts:

• Magnitude of displacement ≤ distance travelled.
• They are equal only if the object moves in one direction without turning back.
• Displacement can be zero even when distance is not (a round trip).
• An instant of time is a single clock reading; a time interval is the gap between two readings.

Quantities with only magnitude are scalars (distance, speed); those with magnitude and direction are vectors (displacement, velocity, acceleration).

Worked example

Riya walks 300 m east from her gate to a shop, then walks 120 m back west to a friend's house. Taking east as positive, find the distance travelled and her displacement.

Try it first, then show the solutionSolution

1. Distance travelled $=300+120=420$ m.
2. Final position $=+300-120=+180$ m from the gate.
3. Displacement $=+180$ m, i.e. 180 m east.

Answer: Distance 420 m; displacement 180 m east.

Average speed and average velocity

NCERT section 4.1 Motion in a Straight Line

$\text{average speed}=\frac{\text{total distance travelled}}{\text{time interval}}$

$\text{average velocity }{v}_{av}=\frac{\text{displacement}}{\text{time interval}}=\frac{s}{t}$

• Average speed has no direction. Average velocity has the direction of the displacement (+ or −).
• SI unit of both: **m s${}^{-1}$** (m/s); also km h${}^{-1}$.
• Convert: $1{\text{ km h}}^{-1}=\frac{1000\text{ m}}{3600\text{ s}}=\frac{5}{18}$ m s${}^{-1}$. So 18 km h${}^{-1}$ = 5 m s${}^{-1}$, 36 km h${}^{-1}$ = 10 m s${}^{-1}$, 72 km h${}^{-1}$ = 20 m s${}^{-1}$.
• Uniform motion: equal distances in equal intervals of time (constant speed). Non-uniform motion: unequal distances in equal intervals.
• Average speed and the magnitude of average velocity are equal only when the object moves in one direction.

India's mathematicians used speed = distance ÷ time long ago; the *Ganitakaumudi* (14th century CE) has a problem on two postmen walking towards each other.

The speedometer shows (nearly) the speed at an instant. 'Velocity' in this chapter means velocity at an instant.

Worked example

A cyclist rides 12 km east in 40 minutes, then 4 km west in 20 minutes. Find the average speed and average velocity for the whole hour.

Try it first, then show the solutionSolution

1. Total distance $=12+4=16$ km; total time $=60$ min $=1$ h.
2. Average speed $=\frac{16\text{ km}}{1\text{ h}}=16$ km h${}^{-1}$.
3. Displacement $=12-4=8$ km east.
4. Average velocity $=\frac{8\text{ km}}{1\text{ h}}=8$ km h${}^{-1}$ east.

Answer: Average speed 16 km h${}^{-1}$; average velocity 8 km h${}^{-1}$ towards the east.

Average acceleration and free fall

NCERT section 4.1 Motion in a Straight Line

$a=\frac{\text{change in velocity}}{\text{time interval}}=\frac{v-u}{t_2-t_1}$

where $u$ is the initial velocity and $v$ the final velocity. SI unit: **m s${}^{-2}$**.

• If the object is speeding up, acceleration is in the same direction as velocity (positive when velocity is positive).
• If it is slowing down, acceleration is opposite to velocity (negative). Everyday words like 'deceleration' or 'retardation' mean this.
• Constant acceleration: velocity changes by equal amounts in equal intervals of time.
• An object moving very fast at a steady velocity has zero acceleration. Acceleration depends on how quickly velocity changes, not on how fast the object moves.
• Acceleration also happens when only the direction of velocity changes (circular motion).

Free fall: a dropped object gains 9.8 m s${}^{-1}$ of speed every second (0, 9.8, 19.6, 29.4, 39.2 m s${}^{-1}$ at 0, 1, 2, 3, 4 s). This constant acceleration is the acceleration due to gravity, $g=9.8$ m s${}^{-2}$, directed downwards. You may choose any origin and positive direction, but must keep it the same throughout a problem.

Worked example

A scooter speeds up from 18 km h${}^{-1}$ to 54 km h${}^{-1}$ in 8 s. Later it brakes from 54 km h${}^{-1}$ to rest in 3 s. Find the average acceleration in each case.

Try it first, then show the solutionSolution

1. Convert: $18\times \frac{5}{18}=5$ m s${}^{-1}$; $54\times \frac{5}{18}=15$ m s${}^{-1}$.
2. Speeding up: $a=\frac{15-5}{8}=\frac{10}{8}=1.25$ m s${}^{-2}$ (along the motion).
3. Braking: $a=\frac{0-15}{3}=-5$ m s${}^{-2}$.
4. The minus sign shows the acceleration is opposite to the velocity.

Answer: 1.25 m s${}^{-2}$ while speeding up; −5 m s${}^{-2}$ while braking.

Plotting and reading position-time graphs

NCERT section 4.2 Graphical Representation of Motion

A graph shows how one quantity changes with another; it is not a route map.

Plotting (Activity 4.3): draw the X-axis (time) and Y-axis (position) meeting at the origin; choose scales that use most of the paper (e.g. 5 divisions = 1 s); mark each (time, position) point; join the points.

Reading a position-time graph (motion in one direction):

Shape	Meaning
straight line parallel to the time axis	object at rest
straight sloping line	constant velocity (uniform motion)
curve getting steeper	velocity increasing (accelerated)
steeper straight line	higher velocity

Slope = velocity: pick two points A $(t_1,s_1)$ and B $(t_2,s_2)$ on the line:
$v=\frac{s_2-s_1}{t_2-t_1}$

The slope tells how fast the Y-quantity changes with the X-quantity.

Worked example

A straight position-time graph passes through (2 s, 30 m) and (6 s, 90 m). Find the velocity. What would the graph look like for an object that stays at 40 m?

Try it first, then show the solutionSolution

1. Slope $=\frac{90-30}{6-2}=\frac{60}{4}=15$ m s${}^{-1}$.
2. A straight line means this velocity is constant.
3. An object staying at 40 m gives a horizontal line at 40 m: slope zero, velocity zero.

Answer: 15 m s${}^{-1}$; a horizontal line at 40 m (at rest).

Velocity-time graphs: slope and area

NCERT section 4.2 Graphical Representation of Motion

Shape of velocity-time graph	Meaning
horizontal line	constant velocity, zero acceleration
straight line rising	velocity increasing at constant acceleration
straight line falling	velocity decreasing at constant (negative) acceleration

Slope = acceleration:
$a=\frac{v-u}{t_2-t_1}$

Area between the graph and the time axis = displacement in that time interval.

• Constant velocity: area of a rectangle = velocity × time.
• Constant acceleration: rectangle + triangle (or a trapezium), e.g. from 10 s to 20 s with velocity rising from 5 to 10 m s${}^{-1}$: $5\times 10+\frac{1}{2}\times 10\times 5=75$ m.

Use units on both axes when you read off values; mixing km h${}^{-1}$ with seconds is a common slip.

Worked example

A car's velocity rises steadily from 0 to 20 m s${}^{-1}$ in 10 s and then stays at 20 m s${}^{-1}$ for 5 s. Use the velocity-time graph to find the acceleration in the first part and the total displacement.

Try it first, then show the solutionSolution

1. Slope of first part $=\frac{20-0}{10}=2$ m s${}^{-2}$.
2. Area of triangle (0 to 10 s) $=\frac{1}{2}\times 10\times 20=100$ m.
3. Area of rectangle (10 to 15 s) $=20\times 5=100$ m.
4. Total displacement $=100+100=200$ m.

Answer: Acceleration 2 m s${}^{-2}$; displacement 200 m.

Deriving the kinematic equations by the graphical method

NCERT section 4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

For constant acceleration, take a velocity-time graph that starts at velocity $u$ (point A) when $t=0$ and rises in a straight line to $v$ (point B) at time $t$. Drop a perpendicular from B to the time axis at D, and draw AC parallel to the time axis, meeting BD at C.

First equation (slope): acceleration = slope of AB:
$a=\frac{BC}{AC}=\frac{v-u}{t}\Rightarrow v=u+at$

Second equation (area): displacement = area under AB = area of rectangle OACD + area of triangle ABC:
$s=ut+\frac{1}{2}\times t\times (v-u)$
Since $v-u=at$:
$s=ut+\frac{1}{2}a{t}^2$

Third equation: put $t=\frac{v-u}{a}$ into the second equation and simplify:
$v^2=u^2+2as$

Two more forms (from The Journey Beyond): $s=\frac{1}{2}(u+v)t$ (area of a trapezium) and $s=vt-\frac{1}{2}a{t}^2$.

These equations are valid only when acceleration is constant. Keep signs consistent: if the motion is positive, a slowing object has negative $a$.

Worked example

A car moving at 10 m s${}^{-1}$ accelerates at 2 m s${}^{-2}$ for 5 s. Find its final velocity and displacement, and check with the third equation.

Try it first, then show the solutionSolution

1. $v=u+at=10+2\times 5=20$ m s${}^{-1}$.
2. $s=ut+\frac{1}{2}a{t}^2=10\times 5+\frac{1}{2}\times 2\times 25=50+25=75$ m.
3. Check: $u^2+2as=100+2\times 2\times 75=400=20^2$. ✓

Answer: 20 m s${}^{-1}$ and 75 m.

Stopping distance and road safety

NCERT section 4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

When a driver sees danger, the vehicle travels:
1. a reaction distance at constant speed while the driver reacts: $u\times t_{reaction}$;
2. a braking distance while slowing to rest: from $0=u^2+2as$, $s=\frac{u^2}{2|a|}$.

Stopping distance = reaction distance + braking distance.

Braking distance depends on $u^2$: doubling the speed makes the braking distance four times as long. It also grows on wet roads, with worn tyres or weak brakes (smaller $|a|$). Reaction distance grows if the driver is tired, distracted or using a phone.

That is why safe following gaps must increase with speed, and why vehicle-to-vehicle (V2V) warning systems are being developed.

Worked example

A car moves at 72 km h${}^{-1}$. The driver takes 0.7 s to react, then the brakes give an acceleration of −5 m s${}^{-2}$. Find the stopping distance.

Try it first, then show the solutionSolution

1. $u=72\times \frac{5}{18}=20$ m s${}^{-1}$.
2. Reaction distance $=20\times 0.7=14$ m.
3. Braking distance $=\frac{u^2}{2|a|}=\frac{400}{10}=40$ m.
4. Stopping distance $=14+40=54$ m.

Answer: 54 m.

Uniform circular motion

NCERT section 4.4 Motion in a Plane

Motion in a plane (two dimensions) includes overtaking, a kicked ball and a satellite in orbit. A simple case is circular motion.

• In one full revolution of radius $R$: distance $=2\pi R$ (circumference); displacement $=0$.
• Average speed for one revolution of time $T$: $v_{av}=\frac{2\pi R}{T}$ Average velocity over one revolution is 0.
• Uniform circular motion: moving on a circle at constant speed.
• The velocity at any point is along the tangent to the circle in the direction of motion. Since the direction changes all the time, the velocity changes, so the motion is accelerated even though the speed is constant.
• Think of a polygon track with more and more sides: the runner turns more and more often, until on a circle the direction changes continuously.

Activity 4.5: a marble running round inside a tape ring moves off in a straight line along the tangent when the ring is lifted. It keeps going in the direction it had at that instant.

Examples (approximately uniform): a merry-go-round, a satellite, the tip of a clock hand, a planet round the Sun.

Worked example

A runner completes a circular track of radius 35 m in 44 s at constant speed. Find (i) her speed (ii) her displacement after half a lap. Take $\pi =\frac{22}{7}$.

Try it first, then show the solutionSolution

1. Circumference $=2\times \frac{22}{7}\times 35=220$ m.
2. Speed $=\frac{220}{44}=5$ m s${}^{-1}$.
3. After half a lap she is at the opposite end of a diameter: displacement $=2R=70$ m.

Answer: (i) 5 m s${}^{-1}$ (ii) 70 m across the diameter.

Science in real life: Describing Motion Around Us

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Distance, Displacement, Speed, Velocity and Motion Graphs
• Equations of Motion, Free Fall and Stopping Distance

Describing Motion Around Us: formulas and key facts

Average speed

$\text{average speed}=\frac{\text{total distance}}{\text{time interval}}$

Average velocity

$v_{av}=\frac{s}{t}=\frac{\text{displacement}}{\text{time interval}}$

Average acceleration

$a=\frac{v-u}{t_2-t_1}$

Unit conversion

$1{\text{ km h}}^{-1}=\frac{5}{18}{\text{ m s}}^{-1}$; multiply m s${}^{-1}$ by $\frac{18}{5}$ to get km h${}^{-1}$.

First equation of motion

$v=u+at$

Second equation of motion

$s=ut+\frac{1}{2}a{t}^2$

Third equation of motion

$v^2=u^2+2as$

Average-velocity form

$s=\frac{1}{2}(u+v)t$

Graph rules

$Slopeofposition-time=velocity;slopeofvelocity-time=acceleration;areaundervelocity-time=displacement.$

Free fall

$g=9.8{\text{ m s}}^{-2}$ downwards; dropped from rest: $v=gt$, $s=\frac{1}{2}g{t}^2$.

Uniform circular motion

$v=\frac{2\pi R}{T}$; displacement after one revolution = 0.

Stopping distance

$u\times t_{reaction}+\frac{u^2}{2|a|}$

Describing Motion Around Us: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Describing Motion Around Us

The question patterns to expect, and the method that works for each.

Tricks to remember Describing Motion Around Us

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Five-eighteenths

  km h⁻¹ to m s⁻¹: multiply by 5/18 (going to the smaller number). m s⁻¹ to km h⁻¹: multiply by 18/5.

  Use it when: Unit conversion in every numerical.

• Slope speaks, area adds

  The slope of a graph gives the rate (velocity or acceleration); the area under a velocity-time graph adds up the displacement.

  Use it when: Graph questions.

• SUVAT checklist

  Write s, u, v, a, t down the margin. The missing one tells you the equation: no $s$ → $v=u+at$; no $v$ → $s=ut+\frac{1}{2}a{t}^2$; no $t$ → $v^2=u^2+2as$.

  Use it when: Choosing the right equation of motion.

• Round trip = zero

  Back where you started? Displacement and average velocity are zero, however far you went.

  Use it when: Trick MCQs on distance and displacement.

• Double speed, four times the skid

  Braking distance goes with $u^2$.

  Use it when: Road-safety and stopping-distance questions.

Describing Motion Around Us: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Describing Motion Around Us

The slips examiners see most often, and what to write instead.

• ✗Mistake: Not converting km h⁻¹ to m s⁻¹

  The data mixes units and students plug them straight in.

  Wrong working

  $a=\frac{54-36}{10}=1.8$ m s${}^{-2}$

  Right working

  $54$ km h${}^{-1}$ $=15$ m s${}^{-1}$, $36$ km h${}^{-1}$ $=10$ m s${}^{-1}$; $a=\frac{15-10}{10}=0.5$ m s${}^{-2}$.

  ✓Do this: Convert every quantity to SI units before substituting.

  Usually costs: The answer mark and often the method mark.

• ✗Mistake: Dropping the minus sign for retardation

  Students think acceleration is always positive.

  Wrong working

  Car stops from 20 m s${}^{-1}$ in 4 s, so $a=5$ m s${}^{-2}$.

  Right working

  $a=\frac{0-20}{4}=-5$ m s${}^{-2}$, i.e. 5 m s${}^{-2}$ opposite to the motion.

  ✓Do this: Always compute $v-u$ in that order and keep the sign.

  Usually costs: 1 mark, and wrong later answers that use $a$.

• ✗Mistake: Treating distance and displacement as the same

  In daily speech they mean the same thing.

  Wrong working

  A swimmer swims one length of a 25 m pool and back: displacement 50 m.

  Right working

  Distance 50 m, displacement 0 m (she is back at the start).

  ✓Do this: Ask: where did it start, where did it end?

  Usually costs: 1 to 2 marks.

• ✗Mistake: Averaging speeds instead of using total distance over total time

  It looks quicker.

  Wrong working

  60 km at 30 km h${}^{-1}$ then 60 km at 60 km h${}^{-1}$: average $=\frac{30+60}{2}=45$ km h${}^{-1}$.

  Right working

  Times are 2 h and 1 h; average speed $=\frac{120}{3}=40$ km h${}^{-1}$.

  ✓Do this: Average speed = total distance ÷ total time, always.

  Usually costs: The full numerical.

• ✗Mistake: Saying uniform circular motion has no acceleration

  The speed is constant.

  Wrong working

  Speed is constant, so velocity and acceleration are unchanged.

  Right working

  Direction changes at every point, so velocity changes and the motion is accelerated.

  ✓Do this: Velocity has direction; changing direction means changing velocity.

  Usually costs: 1 mark in MCQ or assertion-reason.

NCERT exercise guide: Describing Motion Around Us

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Describing Motion Around Us

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The SI unit of velocity is
   (a) m s⁻²
   (b) m s⁻¹
   (c) km h⁻¹
   (d) m

   Hint

   Displacement over time.

   Answer

   (b) m s⁻¹

2. Q2 · 1 mark · MCQExam level

   The slope of a velocity-time graph gives
   (a) displacement
   (b) speed
   (c) acceleration
   (d) distance

   Hint

   Rate of change of velocity.

   Answer

   (c) acceleration

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): In uniform circular motion the average velocity over one full revolution is zero.
   Reason (R): The speed in uniform circular motion is zero.
   (a) A and R are both true, and R explains A correctly.
   (b) A and R are both true, but R does not explain A.
   (c) A is true, R is false.
   (d) A is false, R is true.

   Hint

   Where does it end after one revolution?

   Answer

   (c). Displacement is zero after a full revolution, so average velocity is zero; but the speed is constant and not zero.

4. Q4 · 2 marks · Very short answerExam level

   A lift goes from the 2nd floor up to the 7th floor and then down to the ground floor. Each floor is 3.5 m high. Find the distance travelled by the lift and its displacement from the 2nd floor.

   Hint

   Up 5 floors, down 7 floors.

   Answer

   Up $5\times 3.5=17.5$ m, down $7\times 3.5=24.5$ m: distance $=42$ m. It ends 2 floors below the start: displacement $=7$ m downwards.

5. Q5 · 2 marks · Very short answerHarder exam level

   A train leaves a station from rest with uniform acceleration and reaches 54 km h⁻¹ in 25 s. Find the acceleration and the distance covered in this time.

   Hint

   Convert to m s⁻¹, then use $v=u+at$ and $s=\frac{1}{2}(u+v)t$.

   Answer

   $v=15$ m s⁻¹; $a=\frac{15}{25}=0.6$ m s⁻²; $s=\frac{1}{2}\times 15\times 25=187.5$ m.

6. Q6 · 3 marks · Short answerHarder exam level

   A car moving at 72 km h⁻¹ slows uniformly to 18 km h⁻¹ over 150 m on entering a town. Find the acceleration and the time taken.

   Hint

   20 m s⁻¹ to 5 m s⁻¹; use $v^2=u^2+2as$.

   Answer

   $25=400+2a\times 150\Rightarrow a=-1.25$ m s⁻²; $t=\frac{5-20}{-1.25}=12$ s.

7. Q7 · 3 marks · Short answerHarder exam level

   A ball is dropped from rest and reaches the ground in 2 s. Find the height and the speed just before landing. How long would it take to fall from four times the height?

   Hint

   $s\propto t^2$ from rest.

   Answer

   $h=\frac{1}{2}\times 9.8\times 4=19.6$ m; $v=9.8\times 2=19.6$ m s⁻¹. From $4h$: $t^2$ becomes 4 times, so $t=4$ s.

8. Q8 · 3 marks · Short answerHarder exam level

   Derive $v^2=u^2+2as$ from the first two equations of motion.

   Hint

   Eliminate $t$.

   Answer

   From $v=u+at$, $t=\frac{v-u}{a}$. Substitute in $s=ut+\frac{1}{2}a{t}^2$: $s=\frac{u(v-u)}{a}+\frac{(v-u{)}^2}{2a}=\frac{2uv-2u^2+v^2-2uv+u^2}{2a}=\frac{v^2-u^2}{2a}$, so $v^2=u^2+2as$.

9. Q9 · 3 marks · Short answerHarder exam level

   Data for a trolley rolling down a ramp: at 0, 1, 2, 3 s its distance from the top is 0, 0.2, 0.8, 1.8 m. Is the motion uniform? Find the acceleration assuming it starts from rest.

   Hint

   Distances in successive seconds; then $s=\frac{1}{2}a{t}^2$.

   Answer

   Distances in successive seconds are 0.2, 0.6, 1.0 m (increasing), so it is non-uniform (accelerated). $1.8=\frac{1}{2}a(3{)}^2\Rightarrow a=0.4$ m s⁻² (also $0.2=\frac{1}{2}a(1{)}^2$ gives 0.4).

10. Q10 · 3 marks · Short answerHarder exam level

    Sketch velocity-time graphs for (i) a car at constant velocity (ii) a ball dropped from rest (iii) a car braking uniformly to rest. State what the slope means in each.

    Hint

    Horizontal, rising, falling.

    Answer

    (i) Horizontal line, slope zero, no acceleration. (ii) Straight line from origin rising, slope $g=9.8$ m s⁻². (iii) Straight line falling to the time axis, negative slope = retardation.

11. Q11 · 3 marks · Short answerHOTS / competency

    A student says, 'A book on my desk is at rest.' Another says, 'No, it is moving at about 30 km s⁻¹ with the Earth round the Sun.' Taking the Earth's orbit as a circle of radius $1.5\times 10^8$ km covered in 365 days, check the second claim and explain how both can be right.

    Hint

    Use $v=\frac{2\pi R}{T}$ and think about the reference point.

    Answer

    $v=\frac{2\pi \times 1.5\times 10^8\text{ km}}{365\times 24\times 3600\text{ s}}\approx 29.9$ km s⁻¹, so about 30 km s⁻¹. Both are right: rest and motion depend on the reference point (the desk or the Sun).

12. Q12 · 3 marks · Short answerHOTS / competency

    Two cars start from rest together. Car P accelerates at 2 m s⁻² and car Q at 3 m s⁻². How far apart are they after 10 s? How long until Q is 100 m ahead of P?

    Hint

    Gap $=\frac{1}{2}(a_Q-a_P)t^2$.

    Answer

    $s_P=100$ m, $s_Q=150$ m, so 50 m apart. Gap $=\frac{1}{2}(1)t^2=100\Rightarrow t^2=200$, $t\approx 14.1$ s.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define displacement.Show answer

   The net change in position of an object between two instants, with magnitude and direction.

2. 2.When are distance travelled and magnitude of displacement equal?Show answer

   When the object moves in a straight line in one direction without turning back.

3. 3.What is the SI unit of acceleration?Show answer

   m s⁻² (metre per second squared).

4. 4.Convert 90 km h⁻¹ to m s⁻¹.Show answer

   $90\times \frac{5}{18}=25$ m s⁻¹.

5. 5.What does the slope of a position-time graph give?Show answer

   The velocity.

6. 6.What does the area under a velocity-time graph give?Show answer

   The displacement.

7. 7.What is the value of g?Show answer

   9.8 m s⁻², directed towards the Earth.

8. 8.Write the three equations of motion.Show answer

   $v=u+at$; $s=ut+\frac{1}{2}a{t}^2$; $v^2=u^2+2as$.

9. 9.What shape is the position-time graph of a body at rest?Show answer

   A straight line parallel to the time axis.

10. 10.In which direction is the velocity in uniform circular motion?Show answer

    Along the tangent at that point, pointing the way the object is moving.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find distance and displacement for a round trip?
• ☐Can you convert any speed between km h⁻¹ and m s⁻¹?
• ☐Can you find average speed for a trip with two parts at different speeds?
• ☐Can you give the sign and direction of acceleration when slowing down?
• ☐Can you identify rest, uniform and accelerated motion from position-time graphs?
• ☐Can you find acceleration and displacement from a velocity-time graph?
• ☐Can you derive all three equations of motion graphically?
• ☐Can you choose the right equation using the SUVAT checklist?
• ☐Can you calculate a stopping distance with reaction time?
• ☐Can you explain why uniform circular motion is accelerated?

Describing Motion Around Us: frequently asked questions